# Show that ${\frac{x_1^{n}+x_2^{n}+…+x_n^{n}}{x_1x_2…x_n} + \frac{\sqrt[n]{x_1x_2…x_n}}{x_1+x_2+…+x_n}}\ge{n+\frac{1}{n}}$

Let $n\in\mathbb{N}$ such that ${n}\ge{0}$.

Show that ${\frac{x_1^{n}+x_2^{n}+...+x_n^{n}}{x_1x_2...x_n}+\frac{\sqrt[n]{x_1x_2...x_n}}{x_1+x_2+...+x_n}}\ge{n+\frac{1}{n}}$
when $x_n$ is positive.

I thought I can use AM GM inequality, but if I try to break it in two smaller pieces and apply AMGM I obtained a false affirmation for $\frac{\sqrt[n]{x_1x_2...x_n}}{\ x_1+\ x_2+...+x_n}$. Any help please?

• I think you need $n>0$ and the numerator of second term in LHS must be an $n$th root. Otherwise it is easily shown to be false. Please check. – Macavity Mar 7 '18 at 12:20
• I don't know how to make the numerator $\n$. Please help me. – Septimiu Cristian Mar 7 '18 at 12:54
• Type \sqrt[n] in math mode. – Chris Custer Mar 7 '18 at 13:00
• Thanks. I modified it. – Septimiu Cristian Mar 7 '18 at 13:02

## 1 Answer

Since the inequality is homogeneous you can assume that $x_1x_2...x_n=1.$ Let $x_1^n+x_2^n+...x_n^n = A$ and $x_1+x_2+...x_n = B$ for simplicity. Then, you have some easy inequalities: $$A\geq B\geq n$$ and $A\geq \dfrac{B^n}{n^{n-1}},$ the power mean inequality. Therefore, $$n\left(A+\dfrac{1}{B}\right)\geq \dfrac{B^n}{n^{n-2}}+\dfrac{n}{B}=n^2\left(\dfrac{B^n}{n^{n}}\right)+\dfrac{n}{B}\geq(n^2+1)\left(\dfrac{B^{n^3-1}}{n^{n^3-1}}\right)^{\frac{1}{n^2+1}}\geq n^2+1.$$

This proves your inequality and the equality is attained when all the variables are equal.